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Parametric Curves and Derivatives · Tutorial 911 of 998

Computing Second Derivatives Parametrically

Use the parameter-based second-derivative formula carefully, and check each result against the curve’s derivatives.

Beginner 9 min read

What You'll Learn

  • State when the parametric second-derivative formula applies
  • Differentiate the parametric slope with respect to the parameter
  • Divide by the derivative of x to obtain the second derivative with respect to x
  • Compute second derivatives for polynomial parametrizations
  • Compute second derivatives for trigonometric parametrizations
  • Recognize and correct the mistake of omitting the final division by dx/dt

From Parametric Slope to Second Derivative

A parametrization describes both coordinates of a curve using a parameter \(t\): \(x=x(t)\) and \(y=y(t)\). The parameter records a point on the curve, while derivatives with respect to \(t\) describe how the coordinates change as the parameter changes.

The first derivative \(dy/dx\) measures the curve’s slope, but it is not usually the same as \(dy/dt\). To find the second derivative \(d^2y/dx^2\), we must account for the fact that \(x\) changes as \(t\) changes. The procedure is to differentiate the parametric slope with respect to \(t\), then divide by \(dx/dt\).

Key formula: If \(x(t)\) and \(y(t)\) are twice differentiable and \(x'(t)\ne 0\), then
$$ \frac{d^2y}{dx^2} = \frac{1}{x'(t)}\frac{d}{dt}\left(\frac{y'(t)}{x'(t)}\right). $$
The formula gives the second derivative with respect to \(x\), not with respect to the parameter.

Why the Formula Works

At a parameter value where \(x'(t)\ne 0\), \(x\) changes locally with \(t\), so we can use \(x\) as a local coordinate along the curve. The chain rule then converts differentiation with respect to \(x\) into differentiation with respect to \(t\): for a differentiable quantity \(g(t)\), \(dg/dx=(dg/dt)/(dx/dt)\).

Apply this rule first to \(y\). It gives \(dy/dx=y'/x'\). Apply it again, now to the quantity \(y'/x'\), and the second-derivative formula follows:

1
Write the parametric slope.
By the chain rule, \(dy/dx=y'/x'\), provided \(x'\ne 0\).
2
Differentiate the slope with respect to \(t\).
This produces \(d(y'/x')/dt\), not yet \(d^2y/dx^2\).
3
Convert the derivative back to \(x\).
Divide the result by \(dx/dt=x'\), using the chain rule once more.

The quotient rule gives an equivalent form that is often useful for checking a calculation. Differentiating \(y'/x'\) with respect to \(t\) gives \((x'y''-y'x'')/(x')^2\); dividing by \(x'\) gives:

$$ \frac{d^2y}{dx^2} = \frac{x'(t)y''(t)-y'(t)x''(t)}{[x'(t)]^3}, \qquad x'(t)\ne 0. $$

Both expressions are the same formula. In calculations, it can be clearer to find \(dy/dx\) first, differentiate that expression with respect to \(t\), and then divide by \(x'\). The quotient form provides an independent way to check the result.

Worked Example: A Polynomial Parametrization

Worked Example: A Polynomial Parametrization

Let \(x=t^3+t\) and \(y=t^2\). Find \(d^2y/dx^2\), then check the result at \(t=0\).

First differentiate each coordinate with respect to \(t\):

$$ x'=3t^2+1,\qquad y'=2t. $$

Since \(3t^2+1>0\) for every real \(t\), the required condition \(x'\ne 0\) holds everywhere. The parametric slope is

$$ \frac{dy}{dx} = \frac{y'}{x'} = \frac{2t}{3t^2+1}. $$

Differentiate this quotient with respect to \(t\). The numerator of the quotient-rule result is \(2(3t^2+1)-(2t)(6t)=6t^2+2-12t^2=2-6t^2\), so

$$ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2-6t^2}{(3t^2+1)^2}. $$

This is the rate of change of the slope with respect to \(t\). Divide by \(dx/dt=3t^2+1\) to obtain the rate of change with respect to \(x\):

$$ \frac{d^2y}{dx^2} = \frac{2-6t^2}{(3t^2+1)^3} = \frac{2(1-3t^2)}{(3t^2+1)^3}. $$

For a check at \(t=0\), the expression gives \(2\). The equivalent quotient formula confirms it: \(x'=1\), \(y'=0\), \(x''=0\), and \(y''=2\), so

$$ \frac{x'y''-y'x''}{(x')^3} = \frac{(1)(2)-(0)(0)}{1^3} = 2. $$

The two calculations agree.

Worked Example: Another Polynomial Parametrization

Worked Example: A Polynomial Curve with a Restricted Parameter

Let \(x=t^2+1\) and \(y=t^3-3t\). Find \(d^2y/dx^2\) where the formula applies, and check the value at \(t=1\).

Differentiate the coordinates:

$$ x'=2t,\qquad y'=3t^2-3. $$

The formula requires \(x'=2t\ne 0\), so this calculation applies when \(t\ne 0\). The slope is

$$ \frac{dy}{dx} = \frac{3t^2-3}{2t} = \frac{3}{2}\left(t-\frac{1}{t}\right). $$

Differentiate the slope with respect to \(t\):

$$ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2}\left(1+\frac{1}{t^2}\right). $$

Now divide by \(dx/dt=2t\):

$$ \frac{d^2y}{dx^2} = \frac{\frac{3}{2}(1+t^{-2})}{2t} = \frac{3(t^2+1)}{4t^3}, \qquad t\ne 0. $$

At \(t=1\), this gives \(3(2)/4=3/2\). To check, \(x'=2\), \(y'=0\), \(x''=2\), and \(y''=6\), so the quotient form gives

$$ \frac{x'y''-y'x''}{(x')^3} = \frac{(2)(6)-(0)(2)}{2^3} = \frac{12}{8} = \frac{3}{2}. $$

This agrees with the value from differentiating the parametric slope.

Worked Example: A Trigonometric Parametrization

Worked Example: A Trigonometric Parametrization

Let \(x=2\cos t\) and \(y=3\sin t\). Find \(d^2y/dx^2\) where the formula applies, and check the value at \(t=\pi/6\).

Differentiate both coordinates:

$$ x'=-2\sin t,\qquad y'=3\cos t. $$

Here \(x'\ne 0\) when \(\sin t\ne 0\). At those parameter values, the slope is

$$ \frac{dy}{dx} = \frac{3\cos t}{-2\sin t} = -\frac{3}{2}\cot t. $$

Differentiate with respect to \(t\), using \(d(\cot t)/dt=-\csc^2 t\):

$$ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3}{2}\csc^2 t. $$

Divide by \(dx/dt=-2\sin t\):

$$ \frac{d^2y}{dx^2} = \frac{\frac{3}{2}\csc^2 t}{-2\sin t} = -\frac{3}{4\sin^3 t}, \qquad \sin t\ne 0. $$

At \(t=\pi/6\), \(\sin t=1/2\), so the expression gives \(-3/[4(1/2)^3]=-6\). For a check, at this parameter value \(x'=-1\), \(y'=3\sqrt{3}/2\), \(x''=-\sqrt{3}\), and \(y''=-3/2\). Therefore

$$ \frac{x'y''-y'x''}{(x')^3} = \frac{(-1)(-3/2)-(3\sqrt{3}/2)(-\sqrt{3})}{(-1)^3} = \frac{3/2+9/2}{-1} = -6. $$

The check agrees with the result obtained from the slope.

A Common Mistake: Stopping After Differentiating with Respect to \(t\)

A frequent error is to differentiate \(dy/dx\) with respect to \(t\) and treat the result as \(d^2y/dx^2\). These derivatives use different variables. The chain rule requires one more division by \(dx/dt\).

For example, in the trigonometric parametrization above, differentiating the slope gives \((3/2)\csc^2 t\). A wrong calculation would identify that directly as the second derivative. The correction is to divide by \(x'=-2\sin t\):

$$ \frac{d^2y}{dx^2} = \frac{1}{-2\sin t} \left(\frac{3}{2}\csc^2 t\right) = -\frac{3}{4\sin^3 t}. $$

At \(t=\pi/6\), the incorrect result is \(6\), whereas the correct result is \(-6\). The missing division changes both the value and, in this example, the sign.

Condition to check: The formula requires \(x'(t)\ne 0\) at the parameter value in question, as well as the derivatives used in the calculation. If \(x'(t)=0\), this formula cannot be applied there. That fact alone does not determine whether a second derivative might be defined by another analysis.

Reliable Calculation and Verification

For a parametrization \(x=x(t)\), \(y=y(t)\), a dependable calculation keeps track of which variable is being used at each stage:

  1. Find \(x'\) and \(y'\), and identify the parameter values where \(x'\ne 0\).
  2. Form the slope \(dy/dx=y'/x'\).
  3. Differentiate that slope with respect to \(t\).
  4. Divide the result by \(x'\) to obtain \(d^2y/dx^2\).
  5. Check the result using \((x'y''-y'x'')/(x')^3\), or by substituting a convenient parameter value into both forms.

Writing the intermediate derivative explicitly helps prevent confusion: \(d(y'/x')/dt\) is not yet the requested answer. The final division by \(x'\) converts the rate of change from the parameter to the horizontal coordinate.

The second derivative records how the slope changes as \(x\) changes along the curve. This is why it can differ substantially from the parameter derivative of the slope: the parameter may move through \(x\)-values at different rates. Interpreting the sign or geometric consequences of the second derivative is a separate question; here, the essential task is to compute it with the correct variable conversion.

Check Your Understanding

For each question, keep track of whether the derivative is taken with respect to \(t\) or with respect to \(x\).

  1. State the condition on \(x'(t)\) needed to use the parametric second-derivative formula at a parameter value.
  2. If \(x=t^2+1\) and \(y=t^3\), find \(dy/dx\) and \(d^2y/dx^2\) for \(t\ne 0\).
  3. If \(x=\sin t\) and \(y=\cos t\), find \(d^2y/dx^2\) where the formula applies.
  4. Explain why differentiating \(dy/dx\) with respect to \(t\) does not by itself give \(d^2y/dx^2\).
  5. Use the quotient form of the formula to check a second-derivative calculation at a specific parameter value.

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